Two moles of `NH_(3)` when put into a proviously evacuated vessel (one litre) pertially dissociate into `N_(2)` and `H_(2)`. If at equilibrium one mole of `NH_(3)` is present, the equilibrium constant is

Options

A. `3//4 mol^(2)litre^(-2)`
B. `27//64 mol^(2) litre^(-2)`
C. `27//32 mol^(2) litre^(-2)`
D. `27//16mol^(2) litre^(-2)`

Answer

To find the equilibrium constant for the dissociation of ammonia (NH₃) into nitrogen (N₂) and hydrogen (H₂), we can follow these steps:

### Step 1: Write the balanced chemical equation
The dissociation of ammonia can be represented by the following balanced equation:
\[ 2 \text{NH}_3 (g) \rightleftharpoons \text{N}_2 (g) + 3 \text{H}_2 (g) \]

### Step 2: Set up the initial conditions
Initially, we have:
- Moles of NH₃ = 2 moles
- Moles of N₂ = 0 moles
- Moles of H₂ = 0 moles

Since the volume of the vessel is 1 liter, the initial concentrations are:
- \([NH_3] = 2 \, \text{mol/L}\)
- \([N_2] = 0 \, \text{mol/L}\)
- \([H_2] = 0 \, \text{mol/L}\)

### Step 3: Define the change in moles at equilibrium
Let \(x\) be the number of moles of NH₃ that dissociate. According to the stoichiometry of the reaction:
- Moles of NH₃ at equilibrium = \(2 - x\)
- Moles of N₂ at equilibrium = \(\frac{x}{2}\) (since 1 mole of N₂ is produced for every 2 moles of NH₃ that dissociate)
- Moles of H₂ at equilibrium = \(\frac{3x}{2}\) (since 3 moles of H₂ are produced for every 2 moles of NH₃ that dissociate)

### Step 4: Use the information given in the problem
At equilibrium, it is given that 1 mole of NH₃ is present:
\[ 2 - x = 1 \]
Solving for \(x\):
\[ x = 2 - 1 = 1 \]

### Step 5: Calculate the equilibrium concentrations
Now substituting \(x = 1\) into the expressions for moles at equilibrium:
- Moles of NH₃ = \(2 - 1 = 1\)
- Moles of N₂ = \(\frac{1}{2} = 0.5\)
- Moles of H₂ = \(\frac{3 \times 1}{2} = 1.5\)

Since the volume of the vessel is 1 liter, the concentrations are:
- \([NH_3] = 1 \, \text{mol/L}\)
- \([N_2] = 0.5 \, \text{mol/L}\)
- \([H_2] = 1.5 \, \text{mol/L}\)

### Step 6: Write the expression for the equilibrium constant \(K_c\)
The equilibrium constant \(K_c\) for the reaction is given by:
\[ K_c = \frac{[\text{N}_2][\text{H}_2]^3}{[\text{NH}_3]^2} \]

### Step 7: Substitute the equilibrium concentrations into the expression
Substituting the values we found:
\[ K_c = \frac{(0.5)(1.5)^3}{(1)^2} \]
Calculating \( (1.5)^3 \):
\[ (1.5)^3 = 3.375 \]
Thus,
\[ K_c = \frac{(0.5)(3.375)}{1} = 1.6875 \]

### Step 8: Final Calculation
Now we can simplify:
\[ K_c = \frac{3.375}{2} = 1.6875 \]

### Step 9: Convert to a fraction
To express \(K_c\) in a more standard form:
\[ K_c = \frac{27}{16} \, \text{mol}^2/\text{L}^2 \]

### Final Answer
The equilibrium constant \(K_c\) is:
\[ K_c = \frac{27}{16} \, \text{mol}^2/\text{L}^2 \]

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